On Fri, Oct 28, 2016 at 10:29 AM, Dan Asimov <dasimov@earthlink.net> wrote:
So far, no one has submitted a correct solution.
The only thin I see missing from Fred's construction is that the sequence x_i might converge to an irrational > 0, rather than converging to 0. But this is easily patched up. Choose x_{i+1} rational in (0, x_i/2) and not in H_{i+1}. This must be possible because H_{i+1} cannot contain all the rationals in this interval, since if it did and was closed, it would have to contain an irrational. Now the x_i are guaranteed to converge to 0 so the {x_i} with 0 added is a grunch that is not contained in any H_i. Andy
—Dan
On Oct 28, 2016, at 7:18 AM, James Propp <jamespropp@gmail.com> wrote:
On 10/27/16, Dan Asimov <asimov@msri.org <mailto:asimov@msri.org> <javascript:;>> wrote:
Puzzle: Define a grunch to be any closed and bounded subset of the real line R consisting entirely of rational numbers.
True or false: There is a countable set of grunches such that every grunch is a subset of one of them.
Prove your answer.
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